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Proof step 7 from proof subarray-split-at (previous step)

# ar = n0 + n ⊦ ar = subarray(ar, 0, n) + subarray(ar, n, n0), n0 = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 8:
⊦ ar1 = ar2 ↔ # ar1 = # ar2 ∧ (∀ n. n < # ar1 → ar1[n] = ar2[n])
Proof step 9:
# ar = n0 + n ⊦ n0 = 0, # ar = #(subarray(ar, 0, n) + subarray(ar, n, n0)) ∧ (∀ n1. n1 < # ar → ar[n1] = (subarray(ar, 0, n) + subarray(ar, n, n0))[n1])

Rule argument

spec: array name: Extension seq: ⊦ ar1 = ar2 ↔ # ar1 = # ar2 ∧ (∀ n. n < # ar1 → ar1[n] = ar2[n]) subst: [ar2, ar1] → [subarray(ar, 0, n) + subarray(ar, n, n0), ar] flags: <all paths>

Additional information

Used terms: subarray(ar, 0, n) + subarray(ar, n, n0), ar